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pantera1 [17]
2 years ago
9

Moving plates form what tyes of boundaries?

Physics
1 answer:
HACTEHA [7]2 years ago
8 0

transform, convergent, divergent is the answer

hope it helps

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PLEASE HELP!!! GIVING BRAINLIEST!! ill also answer questions that you have posted if you answer these correctly!!!! (47pts)
Bingel [31]

\huge\mathfrak\red{answer}

Jake was playing making a paper airplane, after making he kept it on the table and went to have food.

Suddenly his brother saw the plane and threw it in the air, The airplane kept flying for about 3m/s and it hit his mother and due to the force the plane stopped.

{kept in on the table>rest

brother threw the plane>moving

it hit his mother>force that stopped it}

(mark me brainliest if you're satisfied with my answer)

\large\sf\red{thank \: you}

8 0
3 years ago
A merry-go-round with a rotational inertia of 600 kg m2 and a radius of 3.0 m is initially at rest. A 20 kg boy approaches the m
nekit [7.7K]

Answer:

The velocity of the merry-go-round after the boy hops on the merry-go-round is 1.5 m/s

Explanation:

The rotational inertia of the merry-go-round = 600 kg·m²

The radius of the merry-go-round = 3.0 m

The mass of the boy = 20 kg

The speed with which the boy approaches the merry-go-round = 5.0 m/s

F_T \cdot r = I \cdot \alpha  = m \cdot r^2  \cdot \alpha

Where;

F_T = The tangential force

I =  The rotational inertia

m = The mass

α = The angular acceleration

r = The radius of the merry-go-round

For the merry go round, we have;

I_m \cdot \alpha_m  = I_m \cdot \dfrac{v_m}{r \cdot t}

I_m = The rotational inertia of the merry-go-round

\alpha _m = The angular acceleration of the merry-go-round

v _m = The linear velocity of the merry-go-round

t = The time of motion

For the boy, we have;

I_b \cdot \alpha_b  = m_b \cdot r^2  \cdot \dfrac{v_b}{r \cdot t}

Where;

I_b = The rotational inertia of the boy

\alpha _b = The angular acceleration of the boy

v _b = The linear velocity of the boy

t = The time of motion

When the boy jumps on the merry-go-round, we have;

I_m \cdot \dfrac{v_m}{r \cdot t} = m_b \cdot r^2  \cdot \dfrac{v_b}{r \cdot t}

Which gives;

v_m = \dfrac{m_b \cdot r^2  \cdot \dfrac{v_b}{r \cdot t} \cdot r \cdot t}{I_m} = \dfrac{m_b \cdot r^2  \cdot v_b}{I_m}

From which we have;

v_m =  \dfrac{20 \times 3^2  \times 5}{600} =  1.5

The velocity of the merry-go-round, v_m, after the boy hops on the merry-go-round = 1.5 m/s.

5 0
3 years ago
What instrument is used to measure force?
katen-ka-za [31]
Barometer duhhhh what’s else a ruler
8 0
3 years ago
Which one of the following acts is an example of body language?
just olya [345]
   An example of body language would be rolling your eyes. That's saying that you're bored, don't care, stop talking to other people without even opening your mouth.
4 0
3 years ago
Read 2 more answers
5. A man in a car is listening to the radio. The radio station is broadcasting at a frequency of 85 MHz from two radio transmitt
nasty-shy [4]

Answer:

Speed = 0.00392 m/s

Explanation:

Solution:

Frequency of the radio = 85 MHz

If we have the frequency, we can calculate the wavelength of the radio wave.

As we know,

Frequency = speed of light/wavelength

wavelength = c/f

c = speed of light = 3 x 10^{8} m/s

So,

Wavelength =  3 x 10^{8} m/s / 85 x 10^{6} Hz

Wavelength = 3.5294 m

Man gets disturbed reception at t = 15 min

t = 15 x 60 = 900 s

t = 900 s

Speed = distance/time

Here, distance is wavelength. So,

Speed =  3.5294 m / 900 s

Speed = 0.00392 m/s

Hence, the man's car is going with speed of 0.00392 m/s

3 0
3 years ago
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